Question: Rewrite the function by completing the square. $f(x)= x^{2} -9 x +14$ $f(x)=$
Explanation: $\begin{aligned} f(x)&= x^2 -9 x +14 \\\\ &= \left(x^2 -9 x\right) +14 \end{aligned}$ Now we want to complete $x^2 -9 x$ into a perfect square. To do that, we should add $\left(\dfrac{{-9}}{2}\right)^2={\dfrac{81}{4}}$ to it: $x^2{-9}x+{\dfrac{81}{4}}=\left(x -\dfrac{9}{2}\right)^2$ We add ${\dfrac{81}{4}}$ inside the parentheses, and subtract ${1}\cdot{\dfrac{81}{4}}$ outside them, to keep the expression equivalent. $\begin{aligned} &\phantom{=} \left(x^2 -9 x\right) +14 \\\\ &=\left(x^2 -9 x+{\dfrac{81}{4}}\right) +14 -{1}\cdot{\dfrac{81}{4}} \\\\ &= \left(x -\dfrac{9}{2}\right)^2 +14 -\dfrac{81}{4} \\\\ &= \left(x -\dfrac{9}{2}\right)^2 -\dfrac{25}{4} \end{aligned}$ In conclusion, the function after completing the square is written as: $f(x)= \left(x -\dfrac{9}{2}\right)^2 -\dfrac{25}{4}$